3.169 \(\int \frac{(a^2+2 a b x+b^2 x^2)^{5/2}}{x} \, dx\)

Optimal. Leaf size=221 \[ \frac{5 a^4 b x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a^3 b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{10 a^2 b^3 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{5 a b^4 x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac{b^5 x^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac{a^5 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

[Out]

(5*a^4*b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (5*a^3*b^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x)
+ (10*a^2*b^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (5*a*b^4*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
4*(a + b*x)) + (b^5*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x)) + (a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[
x])/(a + b*x)

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Rubi [A]  time = 0.0493987, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {646, 43} \[ \frac{5 a^4 b x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a^3 b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{10 a^2 b^3 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{5 a b^4 x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac{b^5 x^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac{a^5 \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x,x]

[Out]

(5*a^4*b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (5*a^3*b^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x)
+ (10*a^2*b^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (5*a*b^4*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
4*(a + b*x)) + (b^5*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x)) + (a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[
x])/(a + b*x)

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^5}{x} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (5 a^4 b^6+\frac{a^5 b^5}{x}+10 a^3 b^7 x+10 a^2 b^8 x^2+5 a b^9 x^3+b^{10} x^4\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac{5 a^4 b x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a^3 b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{10 a^2 b^3 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{5 a b^4 x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac{b^5 x^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac{a^5 \sqrt{a^2+2 a b x+b^2 x^2} \log (x)}{a+b x}\\ \end{align*}

Mathematica [A]  time = 0.0212908, size = 74, normalized size = 0.33 \[ \frac{\sqrt{(a+b x)^2} \left (b x \left (200 a^2 b^2 x^2+300 a^3 b x+300 a^4+75 a b^3 x^3+12 b^4 x^4\right )+60 a^5 \log (x)\right )}{60 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x,x]

[Out]

(Sqrt[(a + b*x)^2]*(b*x*(300*a^4 + 300*a^3*b*x + 200*a^2*b^2*x^2 + 75*a*b^3*x^3 + 12*b^4*x^4) + 60*a^5*Log[x])
)/(60*(a + b*x))

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Maple [A]  time = 0.227, size = 73, normalized size = 0.3 \begin{align*}{\frac{12\,{b}^{5}{x}^{5}+75\,a{b}^{4}{x}^{4}+200\,{a}^{2}{b}^{3}{x}^{3}+300\,{a}^{3}{b}^{2}{x}^{2}+60\,{a}^{5}\ln \left ( x \right ) +300\,{a}^{4}bx}{60\, \left ( bx+a \right ) ^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(5/2)/x,x)

[Out]

1/60*((b*x+a)^2)^(5/2)*(12*b^5*x^5+75*a*b^4*x^4+200*a^2*b^3*x^3+300*a^3*b^2*x^2+60*a^5*ln(x)+300*a^4*b*x)/(b*x
+a)^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65914, size = 120, normalized size = 0.54 \begin{align*} \frac{1}{5} \, b^{5} x^{5} + \frac{5}{4} \, a b^{4} x^{4} + \frac{10}{3} \, a^{2} b^{3} x^{3} + 5 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x + a^{5} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x,x, algorithm="fricas")

[Out]

1/5*b^5*x^5 + 5/4*a*b^4*x^4 + 10/3*a^2*b^3*x^3 + 5*a^3*b^2*x^2 + 5*a^4*b*x + a^5*log(x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2)/x,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/x, x)

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Giac [A]  time = 1.38778, size = 122, normalized size = 0.55 \begin{align*} \frac{1}{5} \, b^{5} x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{4} \, a b^{4} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{10}{3} \, a^{2} b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 5 \, a^{3} b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, a^{4} b x \mathrm{sgn}\left (b x + a\right ) + a^{5} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x,x, algorithm="giac")

[Out]

1/5*b^5*x^5*sgn(b*x + a) + 5/4*a*b^4*x^4*sgn(b*x + a) + 10/3*a^2*b^3*x^3*sgn(b*x + a) + 5*a^3*b^2*x^2*sgn(b*x
+ a) + 5*a^4*b*x*sgn(b*x + a) + a^5*log(abs(x))*sgn(b*x + a)